$f(x, y, z) = (y^2, x^2 - y, 0)$ $\text{curl}(f) = $ $\hat{\imath} + $ $\hat{\jmath} + $ $\hat{k}$
Explanation: $f(x, y, z) = (f_0, f_1, f_2)$ The curl of $f$ : $\begin{aligned} \text{curl}(f) &= \det \begin{bmatrix} {\hat{\imath}} & \hat{\jmath} & \hat{k} \\ \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ \\ f_0 & f_1 & f_2 \end{bmatrix} \\ \\ &= \left( \dfrac{\partial f_2}{\partial y} - \dfrac{\partial f_1}{\partial z} \right) \hat{\imath} \\ \\ &+ \left( \dfrac{\partial f_0}{\partial z} - \dfrac{\partial f_2}{\partial x} \right) \hat{\jmath} \\ \\ &+ \left( \dfrac{\partial f_1}{\partial x} - \dfrac{\partial f_0}{\partial y} \right) \hat{k} \end{aligned}$ $\begin{aligned} f_0(x, y, z) &= y^2 \\ \\ f_1(x, y, z) &= x^2 - y \\ \\ f_2(x, y, z) &= 0 \end{aligned}$ Let's calculate all the partial derivatives we'll need. $f_0$ $f_1$ $f_2$ $\dfrac{\partial}{\partial x}$ $2x$ $0$ $\dfrac{\partial}{\partial y}$ $2y$ $0$ $\dfrac{\partial}{\partial z}$ $0$ $0$ Now we can put it all together. $\begin{aligned} \text{curl}(f) &= \left( \dfrac{\partial f_2}{\partial y} - \dfrac{\partial f_1}{\partial z} \right) \hat{\imath} \\ \\ &+ \left( \dfrac{\partial f_0}{\partial z} - \dfrac{\partial f_2}{\partial x} \right) \hat{\jmath} \\ \\ &+ \left( \dfrac{\partial f_1}{\partial x} - \dfrac{\partial f_0}{\partial y} \right) \hat{k} \\ \\ &= (0 - 0) \hat{\imath} + (0 - 0) \hat{\jmath} + (2x - 2y) \hat{k} \\ \\ &= 0 \hat{\imath} + 0 \hat{\jmath} + (2x - 2y) \hat{k} \end{aligned}$ In conclusion: $\text{curl}(f) = 0 \hat{\imath} + 0 \hat{\jmath} + (2x - 2y) \hat{k}$